Question

the 1st,2nd and 3rd ionization energies of an element are 7ev, 12.5 ev and 42.5 ev respectively. the most stable oxidation state of the element will be???

Answer 1

To remove the first electron, you need to provide 7eV, to remove the second one; you need to access between 12.5eV and 42.5eV that is required to remove the third electron.  

This gives you the most stable oxidation state for this element to be +2.

Here the 3rd and 2nd ionization enthalpy difference is more than 15 eV and evaluate it accordingly.